A vibrating string of length ell under a tens | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a tube closed at one end, the frequencies of harmonics are given by $f_k = \frac{(2k-1)v}{4L}$, where $k=1, 2, 3, \dots$. The first overtone i

Vedclass · Accepted Answer

$344$

Vedclass · Answer

(A) For a tube closed at one end, the frequencies of harmonics are given by $f_k = \frac{(2k-1)v}{4L}$, where $k=1, 2, 3, \dots$. The first overtone is the third harmonic $(k=2)$.Given $L = 0.75 \,m$ and $v = 340 \,m/s$, the frequency of the string $f_s$ is:$f_s = \frac{3 	imes 340}{4 	imes 0.75} = \frac{1020}{3} = 340 \,Hz$.The string produces $4$ beats per second with a tuning fork of frequency $n$, so $n = f_s \pm 4$, which means $n = 340 \pm 4$, so $n = 344 \,Hz$ or $336 \,Hz$.When tension $T$ increases, the frequency of the string $f_s$ increases. Since the beat frequency decreases from $4$ to $2$, the frequency of the string must be approaching the frequency of the tuning fork.If $n = 344 \,Hz$, $f_s$ increases from $340$ towards $344$, reducing the beat frequency ($344 - 340 = 4$ to $344 - 342 = 2$). This is consistent.If $n = 336 \,Hz$, $f_s$ increasing from $340$ would move away from $336$, increasing the beat frequency. Thus, $n = 344 \,Hz$ is correct.

Similar Questions

Vedclass Test Series

Exam Paper Generator

Online Exam Module